Solution to Legendre's Differential Equation

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Theorem

The solution of Legendre's differential equation:

$\displaystyle \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - 2 x \frac {\d y} {\d x} + p \paren {p + 1} y = 0$

can be obtained by Power Series Solution Method.


Proof

Let:

$\displaystyle y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$

such that:

$a_0 \ne 0$

Differentating with respect to $x$:

$\displaystyle \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$
$\displaystyle \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$


Substituting in the original equation:

\(\displaystyle \paren {1 - x^2} \ddot y - 2 x \dot y + p \paren {p + 1} y\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {1 - x^2} \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - x^2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\) \(\) \(\displaystyle \)
\(\displaystyle {} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\)


The summations are dependent upon $n$ and not $x$.

Therefore it is a valid operation to multiply the $x$'s into the summations, thus:

\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\) \(\) \(\displaystyle \)
\(\displaystyle {} - \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n} - 2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}\) \(\) \(\displaystyle \)
\(\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n - 1} - 2 \paren {k - n} }\) \(=\) \(\displaystyle 0\)


Increasing the summation variable $n$ to $n + 2$:

\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}\) \(\) \(\displaystyle \)
\(\displaystyle {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\) \(=\) \(\displaystyle 0\)


Taking the first 2 terms of the second summation out:

\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}\) \(\) \(\displaystyle \)
\(\displaystyle {} + a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }\) \(\) \(\displaystyle \)
\(\displaystyle {} + \sum_{n \mathop = 2}^\infty a_{n - 2} x^{k - n} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {\paren {p + 1} - k \paren {k - 1} }\) \(\) \(\displaystyle \)
\(\displaystyle {} + \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }\) \(=\) \(\displaystyle 0\)


Equating each term to $0$:

\(\text {(1)}: \quad\) \(\displaystyle a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} }\) \(=\) \(\displaystyle 0\)
\(\text {(2)}: \quad\) \(\displaystyle a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }\) \(=\) \(\displaystyle 0\)
\(\text {(3)}: \quad\) \(\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }\) \(=\) \(\displaystyle 0\)


Take equation $(1)$:

$a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } = 0$

It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$.

Thus:

\(\displaystyle p \paren {p + 1} - k \paren {k + 1}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle p^2 - k^2 + p - k\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {p - k} \paren {p + k} + \paren {p - k}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {p - k} \paren {p + k + 1}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle p - k\) \(=\) \(\displaystyle 0\)
\(\, \displaystyle \lor \, \) \(\displaystyle p + k + 1\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle k\) \(=\) \(\displaystyle p\)
\(\, \displaystyle \lor \, \) \(\displaystyle k\) \(=\) \(\displaystyle p - 1\)


Take equation $(2)$:

$a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} } = 0$

As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$.

Thus:

\(\displaystyle a_1 \paren {p \paren {p + 1} - k \paren {k - 1} }\) \(=\) \(\displaystyle 0\)
\(\displaystyle a_1 k\) \(=\) \(\displaystyle 0\) substituting the value of $k$ from $(1)$
\(\displaystyle a_1\) \(=\) \(\displaystyle 0\) as $2 k \ne 0$


Take equation $(3)$:

$\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$

As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.

Thus:

\(\displaystyle 0\) \(=\) \(\displaystyle a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\paren {k - n + 2} \paren {k - n + 1} } {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } a_{n - 2}\)


Since Legendre's differential equation is a second order ODE, it has two independent solutions.


Solution 1 (for $k = p$)

\(\displaystyle y\) \(=\) \(\displaystyle x^p \sum_{n \mathop = 0}^\infty a_n x^{-n}\)
\(\displaystyle \) \(=\) \(\displaystyle x^k \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\)


\(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\paren {p - n + 2} \paren {p - n + 1} } {p \paren {p + 1} - \paren {p - n} \paren {p - n + 1} } a_{n - 2}\) from equation $(4)$
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\paren {p - n + 2} \paren {p - n + 1} } {n \paren {2 p - n + 1} } a_{n - 2}\)


Thus:

\(\displaystyle a_2\) \(=\) \(\displaystyle -\frac {p \paren {p - 1} } {2 \paren {2 p - 1} } a_0\) for $n = 2$
\(\displaystyle a_4\) \(=\) \(\displaystyle -\frac {\paren {p - 2} \paren {p - 3} } {4 \paren {2 p - 3} } a_2\) for $n = 4$
\(\displaystyle \) \(=\) \(\displaystyle \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } a_0\)
\(\displaystyle a_6\) \(=\) \(\displaystyle -\frac {\paren {p - 4} \paren {p - 5} } {6 \paren {2 p - 5} } a_4\) for $n = 6$
\(\displaystyle \) \(=\) \(\displaystyle -\frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \paren {p - 4} \paren {p - 5} } {2 \cdot 4 \cdot 6 \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} } a_0\)


Similarly:

\(\displaystyle a_{2 n}\) \(=\) \(\displaystyle \paren {-1}^n \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \dotsm \paren {p - 2 n - 1} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} \dotsm \paren {2 p - 2 n + 1} } a_0\)


Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$


Substituting for $a_n$:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\right.\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\right)\)

$\Box$


Solution 2 (for $k = p - 1$)

\(\displaystyle y\) \(=\) \(\displaystyle x^{-p - 1} \sum_{n \mathop = 0}^\infty a_n x^{-n}\)
\(\displaystyle \) \(=\) \(\displaystyle x^{-k - 1} \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}\)



\(\displaystyle a_n\) \(=\) \(\displaystyle -\frac {\paren {-p - 1 - n + 2} \paren {-p - 1 - n + 1} } {p \paren {p + 1} - \paren {- p - 1 - n} \paren {-p - n} } a_{n - 2}\) from equation $(4)$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {p + n - 1} \paren {p + n} } {n \paren {2 p + n + 1} } a_{n - 2}\)


Thus:

\(\displaystyle a_2\) \(=\) \(\displaystyle \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } a_0\) for $n = 2$
\(\displaystyle a_4\) \(=\) \(\displaystyle \frac {\paren {p + 3} \paren {p + 4} } {4 \paren {2 p + 5} } a_2\) for $n = 4$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } a_0\)
\(\displaystyle a_6\) \(=\) \(\displaystyle \frac {\paren {p + 5} \paren {p + 6} } {6 \paren {2 p + 7} } a_4\) for $n = 6$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} \paren {p + 5} \paren {p + 6} } {2 \cdot 4 \cdot 6 \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} } a_0\)


Similarly:

\(\displaystyle a_{2 n}\) \(=\) \(\displaystyle \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} \dotsm \paren {2 p + 2 n + 1} } a_0\)


Also:

$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$


Substituting for $a_n$:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\right.\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\right)\)

$\Box$


In summary, the two particular solutions are:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\right.\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\right)\)


and:

\(\displaystyle y\) \(=\) \(\displaystyle a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\right.\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \dotsb\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \left.{\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\right)\)


$\blacksquare$


Also see

The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.


Source of Name

This entry was named for Adrien-Marie Legendre.


Sources